Midpoint & distance

Midpoint & DistanceBy L.D.

Table of ContentsSlide 3: Distance FormulaSlide 4: Midpoint FormulaSlide 5: Find the distance and the midpoint

between the points (-3, 1) and (2, 3).Slide 6: Mini LessonSlide 13: Find the distance and the

midpoint between the points (-2, 1) and (2, 5).

Slide 17: The distance of (4, a) and (1, 6) is 5 units, find all possible values of a.

Slide 23: Alternate Way To Find Distance

Distance FormulaThe distance between (x1, y1) and

(y2, x2) is:d = (x2 – x1)2 + (y2 – y1)2

Note: An alternate way to find distance is on slide 18.

Midpoint FormulaThe midpoint between (x1, y1) and

(y2, x2) is:

( )x1 + x2 y1 + y2

22 ,

Problem 1Find the distance and the


Mini Lesson

When you are placing the problems in formulas like (x1, y1) and (y2, x2) and you need to place your numbers ((-3, 1) and (2, 3)) in the problem, it doesn’t matter which of the choices is x1and y1 or x2and y2, all that matters is that once the designated 1 or 2 from the number is chosen, it stays that way.



First we will find the distance formula, remembering what I said in the mini lesson, do you think you can make it?

Problem 1d = (2 – -3)2 + (3 – 1)2

Now that that is set up with the first point (-3, 1) being (x1, y1) and the second (2, 3) being (x2, y2).

Now I will finish solving the distance formula for the problem on the next slide.

Problem 1

d = (2 – -3)2 + (3 – 1)2

d = (2 +3)2 + (3 – 1)2

d = (5)2 + (2)2

d = 25 + 4d = 29

The final distance between the two points is 29 .



Now we will find the midpoints. Can you try to solve it alone before flipping to the next slide? When I do my problem, (-3, 1) will be equal to (x1, y1) and (2, 3) will be equal to (x2, y2).

Problem 1

( -3 + 2 1 + 3

22 ),( -1 4

22 ),( -1

2 ), 2The final midpoint between the points

is (-1/2, 2).

Problem 1

The distance between (-3, 1) and (2, 3) is 29 , while the midpoint is (-1/2, 2).



Note: In this problem I will treat (-2, 1) as (x1, y1) and (2, 5) as (x2, y2) in midpoint, but vice

versa in distance.

Problem 2First I will find the distance.d = (-2 – 2)2 + (5 – 1)2 d = (-4)2 + (4)2

d = 16 + 16d = 32

The final distance is 32 .

Problem 2Now we have to find the midpoint.

The midpoint is (0, 3)

-2 + 2 1 + 5

22

( 0 6

22 ),( ),

( ), 30

Problem 2The midpoint for this is (0, 3) and

the distance is radical 32.

Bonus

The distance of (4, a) and (1, 6) is 5 units, find all possible values of a.

Bonus

The distance of (4, a) and (1, 6) is 5 units, find all possible values of a.

The first thing to do is to find the “distance” between the two points, disregarding the 5.

Bonus

d = (1 – 4)2 + (6 – a)2

d = (-3)2 + ((6 – a)(6 – a))d = 9 + (36 + a2 – 6a – 6a)d = 9 + 36 + a2 – 12ad = 45 + a2 – 12a

Bonus

Now that we know thatd = 45 + a2 – 12a We can finally use the 5.5 = 45 + a2 – 12a

Bonus

The next step is to square both sides to get rid of the square root sign.(5)2 =( 45 + a2 – 12a )2

25 = 45 + a2 – 12a Next we need to move the 25 so we can put the problem in a format that can be factored.25 = 45 + a2 – 12a -25 -250 = 20 + a2 – 12a

Bonus

Lastly we need to factor it.(If you don’t remember how to factor, go to my blog onto the post titled “Factoring Pt. 1/2 (x^2 + bx + c)”)

0 = a2 – 12a + 20(a – 2)(a – 10)a = 2, a = 10

Alternate Way To Find Distance

To find this, I will being using (-3, 1) and (2, 3) as my example.

The first thing to do when taking this “alternate way” is to graph the two points on a graph.

After that we make a right triangle on the twopoints.

Alternate Way To Find Distance

Now I need to explain why this alternate way works. On the blue line there are 5 spaces, while there are 2 spaces on the green line. If we go back to slide 8 and look at the bold part, you can see that there is a 5 and a 2.

Therefore, once the problem is graphed, the problems must only be added and squared before distance is gotten.

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Midpoint & distance